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Article Properties of Strictly Convex

Dong-Soo Kim 1, Young Ho Kim 2,* and Yoon-Tae Jung 3 1 Department of Mathematics, Chonnam National University, Gwangju 61186, Korea; [email protected] 2 Department of Mathematics, Kyungpook National University, Daegu 41566, Korea 3 Department of Mathematics, Chosun University, Gwangju 61452, Korea; [email protected] * Correspondence: [email protected]; Tel.: +82-53-950-5882

 Received: 28 March 2019; Accepted: 21 April 2019; Published: 29 April 2019 

Abstract: We study functions defined in the E2 in which level curves are strictly convex, and investigate area properties of regions cut off by chords on the level curves. In this paper we give a partial answer to the question: Which has level curves whose lines cut off from a level segment of constant area? In the results, we give some characterization theorems regarding conic sections.

Keywords: ; level curve; ; ; strictly convex ; curvature; equiaffine transformation

1. Introduction The most well-known plane curves are straight lines and , which are characterized as the plane curves with constant Frenet curvature. The next most familiar plane curves might be the conic sections: , and . They are characterized as plane curves with constant affine curvature ([1], p. 4). The conic sections have an interesting area property. For example, consider the following two −1 −1 ellipses given by Xk = g (k) and Xl = g (l) with l > k > 0, where

x2 y2 g(x, y) = + , a, b > 0. a2 b2

For a fixed p on Xk, we denote by A and B the points where the tangent to Xk at p meets Xl. Then the region D bounded by the Xl and the chord AB outside Xk has constant area independent of the point p ∈ Xk. In order to give a proof, consider a transformation T of the plane E2 defined by √ ! b/ ab 0 T = √ . 0 a/ ab √ √ Then Xk and Xl are transformed to concentric circles of radius abk and abl, respectively; the tangent at p to the tangent at the corresponding point p0. Since the transformation T is equiaffine (that is, area preserving), a well-known property of concentric circles completes the proof. For parabolas and hyperbolas given by g(x, y) = y2 − 4ax, a 6= 0 and g(x, y) = x2/a2 − y2/b2, a, b > 0, respectively, it is straightforward to show that they also satisfy the above mentioned area properties. For a proof using 1-parameter group of equiaffine transformations, see [1], pp. 6–7. Conversely, it is reasonable to ask the following question. Question. Are there any other level curves of a function g : R2 → R satisfying the above mentioned area property?

Mathematics 2019, 7, 391; doi:10.3390/math7050391 www.mdpi.com/journal/mathematics Mathematics 2019, 7, 391 2 of 14

A plane curve X in the plane E2 is called ‘convex’ if it bounds a convex domain in the plane E2 [2]. A convex curve in the plane E2 is called ‘strictly convex’ if the curve has positive Frenet curvature κ with respect to the unit N pointing to the convex side. We also say that a convex function f : R → R is ‘strictly convex’ if the graph of f is strictly convex. 2 Consider a smooth function g : R → R. We let Rg denote the of all regular values of the function g. We suppose that there exists an interval Sg ⊂ Rg such that for every k ∈ Sg, the level curve −1 2 Xk = g (k) is a smooth strictly convex curve in the plane E . We let Sg denote the maximal interval in Rg with the above property. If k ∈ Sg, then there exists a maximal interval Ik ⊂ Sg such that each Xk+h with k + h ∈ Ik lies in the convex side of Xk. The maximal interval Ik is of the form (k, a) or (b, k) according to whether the gradient vector ∇g points to the convex side of Xk or not. 2 2 2 2 As examples, consider the two functions gi : R → R, i = 1, 2 defined by gi(x, y) = y + eia x i with positive constant a, ei = (−1) . Then, for the function g1 we have Rg1 = R − {0}, Sg1 = (0, ∞) or (−∞, 0), Ik = (k, ∞) if k > 0, and Ik = (−∞, k) if k < 0. For g2, we get Rg2 = Sg2 = (0, ∞) and

Ik = (0, k) with k ∈ Sg2 . For a fixed point p ∈ Xk with k ∈ Sg and a small h with k + h ∈ Ik, we consider the tangent t to Xk at p ∈ Xk and the closest tangent line ` to Xk+h at a point v ∈ Xk+h, which is to the ∗ tangent line t. We let Ap(k, h) denote the area of the region bounded by Xk and the line ` (See Figure1).

√ √ ∗ 2 2 Figure 1. Ap(1, −3/4) for p = (1, 3/2), v = (1/2, 3/4) and g(x, y) = x /4 + y .

In [3], the following characterization theorem for parabolas was established.

Proposition 1. We consider a strictly convex function f : R → R and the function g : R2 → R given by g(x, y) = y − f (x). Then, the following conditions are equivalent.

∗ 1. For a fixed k ∈ R, Ap(k, h) is a function φk(h) of only h. 2. Up to translations, the function f (x) is a quadratic polynomial given by f (x) = ax2 with a > 0, and hence every level curve Xk of g is a .

In the above proposition, we have Rg = Sg = R and Ik = (k, ∞). In particular, Archimedes proved that every level curve√ Xk (parabola) of the function g(x, y) = 2 2 ∗ y − ax in the Euclidean plane E satisfies Ap(k, h) = ch h for some constant c which depends only on the parabola [4]. In this paper, we investigate the family of strictly convex level curves Xk, k ∈ Sg of a function g : R2 → R which satisfies the following condition. ∗ ∗ (A ): For k ∈ Sg with k + h ∈ Ik, Ap(k, h) with p ∈ Xk is a function φk(h) of only k and h. 2 In order to investigate the family of strictly convex level curves Xk, k ∈ Sg of a function g : R → R satisfying condition (A∗), first of all, in Section2 we introduce a useful lemma which reveals a relation between the curvature of level curves and the gradient of the function g (Lemma3 in Section2). Next, using Lemma3, in Section3 we establish the following characterizations for conic sections. Mathematics 2019, 7, 391 3 of 14

Theorem 1. Let f : R → R be a smooth function. We let g denote the function defined by g(x, y) = ya − f (x), 2 where a is a nonzero with a 6= 1. Suppose that the level curves Xk(k ∈ Sg) of g in the plane E are strictly convex. Then the following conditions are equivalent.

1. The function g satisfies (A∗). 3 2. For k ∈ Sg, κ(p)|∇g(p)| = c(k) is constant on Xk, where κ(p) denotes the curvature of Xk at p ∈ Xk. 3. We have a = 2 and the function f is a . Hence, each Xk is a conic section.

In case the function f (− f , resp.) is itself a non-negative strictly convex function, Theorem1 is a special case (n = 1) of Theorem 2 (Theorem 3, resp.) in [5]. In Section4 we prove the following.

Theorem 2. Let f : R → R be a smooth function. For a rational function j(y) in y, we let g denote the function 2 defined by g(x, y) = f (x) + j(y). Suppose that the level curves Xk(k ∈ Sg) of g in the plane E are strictly convex. Then the following conditions are equivalent.

1. The function g satisfies (A∗). 3 2. For k ∈ Sg, κ(p)|∇g(p)| = c(k) is constant on Xk, where κ(p) denotes the curvature of Xk at p ∈ Xk. 3. Both of the functions j(y) and f (x) are quadratic. Hence, each Xk is a conic section.

When the function g is homogeneous, in Section5 we prove the following characterization theorem for conic sections.

2 Theorem 3. Let g : R → R be a smooth homogeneous function of degree d. Suppose that the level curves Xk 2 of g with k ∈ Sg in the plane E are strictly convex. Then the following conditions are equivalent. 1. The function g satisfies (A∗). 3 2. For k ∈ Sg, κ(p)|∇g(p)| = c(k) is constant on Xk, where κ(p) denotes the curvature of Xk at p ∈ Xk. 3. The function g is given by g(x, y) = (ax2 + 2hxy + by2)d/2,

2 where a, b and h satisfy ab − h 6= 0. Thus, each Xk is either a or an ellipse centered at the origin.

Finally, we prove the following in Section6.

Proposition 2. There exists a function g(x, y) = f (x) + j(y) which satisfies the following.

1. Every level curve of g is strictly convex with Sg = R. 3 2. For k ∈ Sg, κ(p)|∇g(p)| = c(k) is constant on Xk, where κ(p) denotes the curvature of Xk at p ∈ Xk. 3. The function g does not satisfy (A∗).

A lot of properties of conic sections (especially, parabolas) have been proved to be characteristic ones [6–13]. For hyperbolas and ellipses centered at the origin, using the support function h and the curvature function κ of a plane curve, a characterization theorem was established [14], from which we get the proof of Theorem3 in Section5. Some characterization theorems for hyperplanes, circular hypercylinders, hyperspheres, elliptic + and elliptic in the Euclidean space En 1 were established in [5,15–19]. For a n+1 characterization of hyperbolic space in the Minkowski space E1 , we refer to [20]. In this article, all functions are smooth (C(3)).

2. Preliminaries

Suppose that X is a smooth strictly convex curve in the plane E2 with the unit normal N pointing to the convex side. For a fixed point p ∈ X and for a sufficiently small h > 0, we take the line ` passing Mathematics 2019, 7, 391 4 of 14 through the point p + hN(p) which is parallel to the tangent t to X at p. We denote by A and B the points where the line ` meets the curve X and put Lp(h) and Ap(h) the length of the chord AB of X and the area of the region bounded by the curve and the line `, respectively. Without loss of generality, we may take a (x, y) of E2 with the origin p, the tangent line to X at p is the x-axis. Hence X is locally the graph of a strictly convex function f : R → R with f (p) = 0. For a sufficiently small h > 0, we get Z Ap(h) = {h − f (x)}dx, Ip(h) Z Lp(h) = 1dx, Ip(h) where we put Ip(h) = {x ∈ R| f (x) < h} and Lp(h) is nothing but the length of Ip(h). Note that we also have Z h Z h Z Ap(h) = Lp(y)dy = { 1dx}dy, y=0 y=0 Ip(y) from which we obtain 0 Ap(h) = Lp(h). We have the following [3]:

Lemma 1. Suppose that X is a smooth strictly convex curve in the plane E2. Then for a point p ∈ X we have √ 1 2 2 lim √ Lp(t) = p t→0 t κ(p) and √ 1 4 2 lim √ Ap(t) = p , t→0 t t 3 κ(p) where κ(p) is the curvature of X at p.

−1 2 Now, we consider the family of strictly convex level curves Xk = g (k) of a function g : R → R with k ∈ Sg. ∗ Suppose that the function g satisfies condition (A ). For each k ∈ Sg and p ∈ Xk we denote by κ(p) the curvature of Xk at p By considering −g if necessary, we may assume that Ik is of the form (k, a) with k < a, and hence we have N = ∇g/|∇g| on Xk. For a fixed point p ∈ Xk and a small t > 0, we have

∗ Ap(t) = Ap(k, h(t)) = φk(h(t)), where h = h(t) is a function with h(0) = 0. Differentiating with respect to t gives

0 0 0 Lp(t) = Ap(t) = φk(h)h (t),

0 where φk(h) is the derivative of φk with respect to h. This shows that

0 r 1 φk(h) h(t) 0 √ Lp(t) = √ h (t). (1) t h t

Next, we use the following lemma for the limit of h0(t) as t → 0. Mathematics 2019, 7, 391 5 of 14

Lemma 2. We have 0 lim h (t) = |∇g(p)|. (2) t→0

Proof. See the proof of Lemma 8 in [5].  It follows from (2) that r h(t) q lim = |∇g(p)|. (3) t→0 t √ 0 Together with Lemma1, (2) and (3), (1) implies that limh→0 φk(h)/ h exists (say, γ(k)), which is independent of p ∈ Xk. Furthermore, we also obtain

8 κ(p)|∇g(p)|3 = , γ(k)2 which is constant on the level curve Xk. Finally, we obtain the following lemma which is useful in the proof of Theorems stated in Section1.

2 ∗ Lemma 3. We suppose that a function g : R → R satisfies condition (A ). Then, for each k ∈ Sg, on Xk the function defined by κ(p)|∇g(p)|3 = c(k) is constant on Xk, where κ(p) is the curvature of Xk at p.

Remark 1. Lemma3 is a special case (n = 1) of Lemma 8 in [5]. For conveniences, we gave a brief proof.

3. Proof of Theorem 1 In this section, we give a proof of Theorem1 stated in Section1. For a nonzero real number a(6= 1) and a smooth function f : R → R, we investigate the level 2 a curves of the function g = ga : R → R defined by ga(x, y) = y − f (x). Suppose that the function g satisfies condition (A∗). Then, it follows from Lemma3 that on the −1 level curve Xk = g (k) with k ∈ Sg we have

κ(p)|∇g(p)|3 = c(k), (4) where c(k) is a function of k ∈ Sg. a Note that for p = (x, y) ∈ Xk with y = f (x) + k we have

3 0 2 2 2a−2 3 |∇g(p)| = { f (x) + a ( f (x) + k) a } 2 , and hence 2a−2 00 a−2 0 κ(p)|∇g(p)|3 = |a2( f (x) + k) a f (x) + a(1 − a)( f (x) + k) a f (x)2|. (5)

Thus, it follows from (4) and (5) that for some nonzero c = c(k) with k ∈ Sg, the function f (x) satisfies

2a−2 00 a−2 0 a2( f (x) + k) a f (x) + a(1 − a)( f (x) + k) a f (x)2 = c(k), which can be rewritten as

00 1 − a − 0 c(k) 2−2a f (x) + ( f (x) + k) 1 f (x)2 = ( f (x) + k) a . (6) a a2 By differentiating (6) with respect to k, we get

0 0 c (k) 2 c(k) 2−a f (x)2 = ( f (x) + k) a − 2 ( f (x) + k) a . (7) a(a − 1) a2 Mathematics 2019, 7, 391 6 of 14

Putting u = f (x) + k and v = du/dx = f 0(x), we get from (6)

dv 1 − a − c(k) 2−2a − + u 1v = u a v 1, (8) du a a2 which is a Bernoulli . By letting w = v2, we obtain

dw 2 − 2a − 2c(k) 2−2a + u 1w = u a . (9) du a a2

2−2a Since u a is an integrating factor of (9), we get

d 2−2a 2c(k) 4−4a (wu a ) = u a . (10) du a2 Now, in order to integrate (10), we divide by some cases as follows. 4 Case 1. Suppose that a = 3 . Then, from (10) we have

9c(k) √ w = { ln u + b(k)} u, (11) 8 where b = b(k) is a constant. Since u = f (x) + k and w = f 0(x)2,(7) and (11) show that

8 c(k) ln( f (x) + k) + b(k) = 2c0(k)( f (x) + k) − c(k). (12) 9 By differentiating (12) with respect to x, we obtain

2c0(k)( f (x) + k) = c(k). (13)

Since c(k) is nonzero, (13) leads to a contradiction. 4 Case 2. Suppose that a 6= 3 . Then, from (8) we have

2c(k) 2 − a 2a − 2 w = a(k)uα + b(k)uβ, a(k) = , α = , β = , (14) (4 − 3a)a a a where b = b(k) is a constant. Since u = f (x) + k and w = f 0(x)2, it follows from (7) and (14) that

0 3a−4 c (k) a − 2 b(k)( f (x) + k) a = ( f (x) + k) − 4c(k) . (15) a(a − 1) a2(3a − 4)

By differentiating (15) with respect to x, we get

0 2a−4 c (k) b(k)( f (x) + k) a = . (16) a(a − 1)

If b(k) 6= 0, then (16) shows that a = 2. If b(k) = 0, then it follows from (15) and (16) that c0(k) = 0, and hence a = 2. Finally, we consider the remaining case as follows. Case 3. Suppose that a = 2. Then, it follows from (7) that for the constant c = c(k)

c0(k) c(k) f 0(x)2 = ( f (x) + k) − . (17) 2 2 Mathematics 2019, 7, 391 7 of 14

If c0(k) = 0, that is, c is independent of k, then (17) shows that f (x) is a linear function. Hence 2 0 each level curve Xk of the function g(x, y) = y − f (x) is a parabola. If c (k) 6= 0, then differentiating both sides of (17) with respect to x shows

4 f 00(x) = c0(k).

This yields that f (x) is a quadratic function and c(k) is a linear function in k. Combining Cases 1–3, we proved the following:

1) ⇒ 2) ⇒ 3).

Conversely, suppose that the function g is given by

g(x, y) = y2 − (ax2 + bx + c),

2 2 where a, b and c are constants with a + b 6= 0. Then, each level curve Xk of g is an ellipse (a < 0), a hyperbola (a > 0) or a parabola (a = 0, b 6= 0). It follows from Section1 or [ 4], pp. 6–7 that the function g satisfies condition (A∗). This shows that Theorem1 holds.

Remark 2. It follows from the proof of Theorem1 that the constant c = c(k) is independent of k if g(x, y) = y2 − 4ax, a 6= 0 and it is a linear function in k if g(x, y) = y2 − ax2, a 6= 0.

Finally, we note the following.

∗ Remark 3. Suppose that a smooth function g : R2 → R satisfies condition (A ) with

κ(p)|∇g(p)|3 = c(k),

−1 where p ∈ Xk = g (k) and k ∈ Sg. Then for any positive constant d, there exists a composite function G = φ ◦ g satisfying condition (A∗) with

κ(p)|∇G(p)|3 = d. (18)

Note that the function G = φ ◦ g has the same level curves as the function g. In order to prove (18), we denote by φ(t) an indefinite integral of the function (d/c(t))1/3. Then for p ∈ G−1(k) = g−1(φ−1(k)) we get

|∇G(p)| = φ0(g(p))|∇g(p)|.

Hence, on each level curve G−1(k) = g−1(φ−1(k)) we obtain

κ(p)|∇G(p)|3 = c(φ−1(k))φ0(φ−1(k))3 = d.

4. Proof of Theorem 2 In this section, we give a proof of Theorem2. We consider a function g defined by g(x, y) = f (x) + j(y) for some functions f (x) and j(y). −1 Then at the point p ∈ Xk = g (k) we have

3 0 2 0 2 3 |∇g(p)| = { f (x) + j (y) } 2 , κ(p)|∇g(p)|3 = | f 00(x)j0(y)2 + f 0(x)2j00(y)|. Mathematics 2019, 7, 391 8 of 14

Suppose that the function g satisfies condition (A∗). Then, it follows from Lemma3 that on the level curve Xk : f (x) + j(y) = k we get for some nonzero constant c = c(k)

f 00(x)j0(y)2 + f 0(x)2j00(y) = c(k), (19)

0 0 which shows that the set V = {(x, y) ∈ Xk| f (x) = 0 or j (y) = 0} has no interior points in the level curve Xk. Hence by continuity, without loss of generality we may assume that V is empty. First, we consider y as a function of x and k. Then, we rewrite (19) as follows

j00(y) c(k) f 00(x) + f 0(x)2 = , j(y) + f (x) = k. (20) j0(y)2 j0(y)2

Putting u = − f (x) + k and v = du/dx = − f 0(x), we get

dv j00(y) c(k) − v = − v−1, du j0(y)2 j0(y)2 which is a Bernoulli equation. By letting w = v2 = f 0(x)2, we obtain

dw 2j00(y) 2c(k) − w = − . (21) du j0(y)2 j0(y)2

Since u = j(y), we see that j0(y)−2 is an integrating factor of (21). Hence we get

d (wj0(y)−2) = −2c(k)j0(y)−4. du Thus we obtain f 0(x)2 = w = −2c(k)j0(y)2{φ(y) + d(k)}, (22) where φ(y) is a function of y satisfying φ0(y) = j0(y)−3 and d = d(k) is a constant.

On the other hand, by differentiating (20) with respect to k, we get

f 0(x)2{j0(y)j000(y) − 2j00(y)2} = c0(k)j0(y)2 − 2c(k)j00(y). (23)

It follows from (22) and (23) that

a(k)j0(y)2 − j00(y) = j0(y)2{2j00(y)2 − j0(y)j000(y)}{φ(y) + d(k)}, (24)

0 ( ) = c (k) where we use a k 2c(k) . Or equivalently, we get

a(k)j0(y)2 − j00(y) φ(y) + d(k) = , (25) j0(y)2{2j00(y)2 − j0(y)j000(y)} where the denominator does not vanish. Even though j(y) was assumed to be C(3), (24) implies that the function {2j00(y)2 − j0(y)j000(y)} is differentiable. By differentiating (25) with respect to x, it is straightforward to show that

d {a(k)j0(y)2 − j00(y)} {2j00(y)2 − j0(y)j000(y)} = 0. (26) dy

Together with (24), (26) yields that 2j00(y)2 − j0(y)j000(y) is constant. Hence, for some constant α we have 2j00(y)2 − j0(y)j000(y) = α. (27) Mathematics 2019, 7, 391 9 of 14

Next, interchanging the role of x and y in the above discussions, we consider x as a function of y and k. Then, (22) gives j0(y)2 = −2c(k) f 0(x)2{ψ(x) + e(k)}, (28) where ψ(x) is a function of x satisfying ψ0(x) = f 0(x)−3 and e = e(k) is a constant. In the same argument as the above, we obtain the corresponding from (23)–(27). For example, we get from (26) d {a(k) f 0(x)2 − f 00(x)} {2 f 00(x)2 − f 0(x) f 000(x)} = 0. (29) dx Thus, for some constant β, we also get

2 f 00(x)2 − f 0(x) f 000(x) = β. (30)

By integrating (24) and (30) respectively, we obtain for some constants γ and δ

2j00(y)2 = γj0(y)4 + α (31) and its corresponding equation 2 f 00(x)2 = δ f 0(x)4 + β. (32)

Differentiating (19) with respect to x, we have

1 d { f 000(x)j0(y)3 − f 0(x)3j000(y)} = { f 00(x)j0(y)2 + f 0(x)2j00(y)} = 0. (33) j0(y) dx

Together with (31) and (32), this shows that j(y) is quadratic in y if and only if f (x) is quadratic in x. Hereafter, we assume that neither f (x) nor j(y) are quadratic. Then, combining (27), (30), (31) and (32), it follows from (33) that (γ − δ) f 0(x)4j0(y)4 = 0, which shows that γ = δ. Hence, for a nonzero constant γ the functions f (x) and j(y) satisfy, respectively 2 f 00(x)2 = γ f 0(x)4 + β (34) and 2j00(y)2 = γj0(y)4 + α. (35)

Differentiating (34) and (35) with respect to x and y, respectively, implies

f 000(x) = γ f 0(x)3, j000(y) = γj0(y)3, (36) where γ is a nonzero constant. Conversely, we prove the following for later use in Section6.

Lemma 4. Suppose that the functions f (x) and j(y) satisfy (34) and (35) for some constants α and β, 3 respectively. Then on each level curve Xk with k ∈ Sg of the function g(x, y) = f (x) + j(y), κ(p)|∇g(p)| is constant.

Proof. Using (36), it follows from the first equality of (33) that on the level curve Xk of the function g, we have d { f 00(x)j0(y)2 + f 0(x)2j00(y)} = 0. dx This completes the proof of Lemma4.  Finally, we proceed on our way. We divide by two cases as follows. Mathematics 2019, 7, 391 10 of 14

Case 1. Suppose that j(y) is a polynomial of degree deg h = n ≥ 3. Then, by counting the degree of both sides of the second equation in (36) we see that the constant γ must vanish. This contradiction shows that the polynomial j(y) is quadratic. Case 2. Suppose that j(y) is a rational function given by

s(y) j(y) = , q(y) where q and s are relatively prime polynomials of degree deg q = m(≥ 1) and deg s = n(≥ 0), respectively. Subcase 2-1. Suppose that m ≥ n. Then we get from (35) that

αq(y)8 = γA(y)4 − 2B(y)2, (37) where we put

A(y) = s0(y)q(y) − s(y)q0(y), B(y) = A0(y)q(y)2 − 2q(y)q0(y)A(y).

Since the degree of the right hand side of (37) is less than or equal to 8m − 4, (37) shows that α must vanish. By integrating (300) with α = 0, we obtain for some constant a and b

1 j(y) = ln |ay + b|, a which is a contradiction. Subcase 2-2. Suppose that m ≤ n − 2. We put

s(y) t(y) j(y) = = r(y) + , q(y) q(y) where deg r = a = n − m ≥ 2 and deg t ≤ m − 1. Then we get from (300) that

γ{r0(y)q(y)2 + A(y)}4 = 2{r00(y)q(y)4 + B(y)}2 − αq(y)8, (38) where we put

A(y) = t0(y)q(y) − t(y)q0(y), B(y) = A0(y)q(y)2 − 2q(y)q0(y)A(y).

Since the degree of the left hand side of (38) is 8m + 4a − 4 and the degree of the right hand side of (38) is less than or equal to 8m + 2a − 4, we see that γ must vanish, which is a contradiction. Hence this case cannot occur.

Subcase 2-3. Suppose that m = n − 1. Then we have r(y) = r0y + r1 with r0 6= 0,

A(y) j0(y) = r + , A(y) = t0(y)q(y) − t(y)q0(y), 0 q(y)2

B¯(y) j00(y) = , B¯(y) = A0(y)q(y) − 2A(y)q0(y) q(y)3 and C(y) j000(y) = , C(y) = B¯0(y)q(y)3 − 3B¯(y)q(y)2q0(y). q(y)6 Mathematics 2019, 7, 391 11 of 14

It follows from the second equation of (36) that

2 3 γ{r0q(y) + A(y)} = C(y).

Note that the left hand side is of degree 6m, but the right hand side is of degree deg C ≤ 6m − 4. Hence, the constant γ must vanish, which is a contradiction. Thus, this case cannot occur. Combining Cases 1 and 2, we see that the function j(y) is a quadratic polynomial. Therefore, Theorem1 completes the proof of Theorem2.

5. Proof of Theorem 3 In this section, we give a proof of Theorem3. Consider a smooth homogeneous function g : R2 → R of degree d. Suppose that the function g ∗ −1 satisfies (A ). Then, it follows from Lemma3 that on the level curve Xk = g (k) with k ∈ Sg we have

κ(p)|∇g(p)|3 = c(k), (39) where c(k) is a nonzero function of k ∈ Sg. We recall the support function h(p) on the level curve Xk, which is defined by

h(p) = hp, N(p)i , where N(p) denotes the unit normal to Xk. Note that the unit normal N(p) to Xk is given by

∇g(p) N(p) = . |∇g(p)|

Since the function g is homogeneous of degree d, by the Euler identity, on Xk we obtain

hp, ∇g(p)i dk h(p) = = . (40) |∇g(p)| |∇g(p)|

Thus, it follows from (39) and (40) that Xk satisfies

c(k) κ(p) = h(p)3. (dk)3

Now, we use the following characterization theorem [14].

Proposition 3. Suppose that X is a smooth curve in the plane E2 of which curvature κ does not vanish identically. Then X satisfies for some constant c

κ(p) = ch(p)3. if and only if X is a connected open arc of either a hyperbola or an ellipse centered at the origin.

The above proposition shows that for each k ∈ Sg, the level curve Xk is either a hyperbola centered at the origin or an ellipse centered at the origin. Without loss of generality, we may assume that 1 ∈ Sg. −1 Then, the level curve X1 = g (1) is given by

ax2 + 2hxy + by2 = 1, (41) where a, b and h satisfy ab − h2 6= 0. We claim that g(x, y) = (ax2 + 2hxy + by2)d/2. (42) Mathematics 2019, 7, 391 12 of 14 where a, b and h satisfy ab − h2 6= 0. 2 In order to prove (42), for a fixed point p = (x, y) ∈ R we let g(x, y) = k, that is, p = (x, y) ∈ Xk. Then we have for t = k−1/d g(tx, ty) = 1.

Hence we get from (41) ax2 + 2hxy + by2 = t−2 = k2/d.

This shows that g(x, y) = k = (ax2 + 2hxy + by2)d/2, which proves the above mentioned claim. Therefore, the proof of Theorem3 was completed.

6. Proof of Proposition 2 In this section, we prove Proposition2. We denote by ψ(t) the function defined by

1 ψ0(t) = √ , ψ(0) = 0 1 + t4 and we put Z ∞ a = (t4 + 1)−1/2dt. 0 Then, both of ψ : (−∞, ∞) → (−a, a) and ψ−1 : (−a, a) → (−∞, ∞) are strictly increasing odd functions. Now, we consider the function g(x, y) = f (x) + j(y) defined on the domain U = (0, a) × (0, ∞) ⊂ R2 with j(y) = ln y and f (x) = ln ψ−1(x).

Then we have Sg = Rg = R and Ik = (k, ∞). Furthermore, it is straightforward to show that the functions f (x) and j(y) satisfies (34) and (300) respectively, where we put γ = 2, α = 0 and β = −8. 3 Thus, Lemma4 implies that on each level curve Xk of the function g(x, y) = f (x) + j(y), κ(p)|∇g(p)| is constant. ∗ However, we show that the function g cannot satisfy condition (A ) as follows. For each k ∈ Sg = −1 R, the level curve Xk = g (k) of g are given by

yψ−1(x) = ek, x, y > 0.

Note that Xk is the graph of the strictly convex function given by

ek y = , x ∈ (0, a), ψ−1(x) which satisfies dy d2y < 0, > 0 dx dx2 and dy dy lim y = ∞, lim = −∞, lim y = 0, lim = −ek. x→0 x→0 dx x→a x→a dx

Hence, each level curve Xk approaches the point (a, 0) and the y-axis is an of Xk. For a fixed point v of X0 and a negative number h < 0, let p ∈ Xh be the point where the tangent t to Xh is parallel to the tangent ` to X0 at v. We denote by A(h) and B(h) the points where the tangent ` to X0 at v intersects the level curve Xh. Mathematics 2019, 7, 391 13 of 14

Suppose that the function g satisfies condition (A∗). Then, the area of the region enclosed by ∗ Xh and the chord A(h)B(h) of Xh is Ap(h, −h) = φh(−h), which is independent of v. We also denote by A and B the points where the tangent ` to X0 at v meets the coordinate axes, respectively. Then, A(h) and B(h) tend to A and B, respectively, as h tends to −∞. Furthermore, as h tends to −∞, φh(−h) goes to the area of the OAB, where O denotes the origin. Thus, the area of the triangle OAB is independent of the point v ∈ X0. This contradicts the following lemma, which might be well known. Therefore the function g(x, y) = f (x) + j(y) does not satisfy condition (A∗). This gives a proof of Proposition2.

Lemma 5. Suppose that X denotes the graph of a strictly convex function f : I → R defined on an open interval I. Then X satisfies the following condition (A) if and only if X is a part of the hyperbola given by xy = c for some nonzero c.

(A): For a point v ∈ X, we put A and B at the points where the tangent ` to X at v intersects coordinate axes, respectively. Then the area of the triangle OAB is independent of the point v ∈ X.

Proof. Suppose that X satisfies condition (A). Then, f 0(x) vanishes nowhere on the interval I. For a point v = (x, f (x)), the area A(x) of the triangle OAB is given by

−1 A(x) = {x f 0(x) − f (x)}2. (43) 2f 0(x)

Differentiating (43) with respect to x gives

−1 {x2 f 0(x)2 − f (x)2} f 00(x) = 0. (44) 2f 0(x)2

By assumption, f 00(x) > 0. Hence, we get from (44)

x2 f 0(x)2 − f (x)2 = 0, which shows that X is a hyperbola given by xy = c for some nonzero c. It is trivial to prove the converse. 

Remark 4. For some higher dimensional analogues of Lemma5, see [19].

Author Contributions: D.-S.K. and Y.H.K. set up the problem and computed the details and Y.-T.J. checked and polished the draft. Funding: The first named author was supported by the Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education (NRF-2018R1D1A3B05050223). The second named author was supported by the National Research Foundation of Korea (NRF) Grant funded by the Korea Government (MSIP) grant number 2016R1A2B1006974. Acknowledgments: We would like to thank the referee for the careful review and the valuable comments to improve the paper. Conflicts of Interest: The authors declare no conflict of interest.

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